[ 二叉树 ] 删除搜索二叉树中的结点

450. 删除二叉搜索树中的节点 - 力扣（LeetCode） (leetcode-cn.com)

删除搜索二叉树中的结点

递归

• 没有孩子，直接删除
• 没左树，右树补；没右树，左树补
• 左右树都存在，左树补到右树最左结点的左边； 或右树补到左树最右结点的右边

class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if (!root) return nullptr;
		// 寻找删除点
        if (root->val > key) root->left = deleteNode(root->left, key);
        else if (root->val < key) root->right = deleteNode(root->right, key);
        else {
            // 没有孩子，直接删除
            if (!root->left && !root->right) return nullptr;

            // 没左树，右树补；没右树，左树补
            if (!root->left) return root->right;
            if (!root->right) return root->left;

            // 左右树都存在
            // 左树补到右树最左结点的左边
            //TreeNode* node = root->right;
            //while (node->left) node = node->left;
            //node->left = root->left;
            //root = root->right;
            
            // 或右树补到左树最右结点的右边
            TreeNode* node = root->left;
            while (node->right) node = node->right;
            node->right = root->right;
            root = root->left;
        }
        return root;
    }
};

非递归

class Solution {
public:
    // 同递归的删除逻辑
    TreeNode* deleteOneNode(TreeNode* node) {
        if (!node || (!node->right && !node->left)) return nullptr;
        if (!node->left) return node->right;
        if (!node->right) return node->left;
        TreeNode* tmp = node->right;
        while (tmp->left) tmp = tmp->left;
        tmp->left = node->left;
        return node->right;
    }

    TreeNode* deleteNode(TreeNode* root, int key) {
        if (!root) return nullptr;
        // 如果删除的是根结点，没有父结点，则直接返回删除后的结果
        if (root->val == key) return deleteOneNode(root);

        TreeNode* cur = root;
        TreeNode* parent = nullptr;
        while (cur) {
            if (cur->val == key) break;
            parent = cur;
            if (cur->val > key) cur = cur->left;
            else cur = cur->right;
        }
        // 判断删除的点是左子树还是右子树，方便重接
        if (parent->left && parent->left->val == key) parent->left = deleteOneNode(cur);
        if (parent->right && parent->right->val == key) parent->right = deleteOneNode(cur);
        return root;
    }
};